The enolic form of acetone `(CH_(3)-overset(OH)overset(|)C=CH_2)` contains

Options

A. `10 sigma`-bonds, `1pi` bond and 1 lone pair
B. `8sigma`-bonds , `2pi`-bonds and 2 lone pairs
C. `9sigma`bonds , `1pi` bond and 2 lone pairs
D. `9sigma` bonds, `2pi` bonds and 1 lone pair

Answer

To solve the problem regarding the enolic form of acetone, we will follow these steps:

### Step 1: Draw the Structure of Enolic Form of Acetone
The enolic form of acetone can be represented as:
```
H
|
H - C = C - OH
|
H
```
In this structure, we have a carbon-carbon double bond and a hydroxyl group (-OH) attached to one of the carbons.

### Step 2: Identify the Atoms and Bonds
In the enolic form:
- There are three carbon atoms (C).
- There are six hydrogen atoms (H).
- There is one oxygen atom (O).

### Step 3: Count the Sigma Bonds
- Each single bond is a sigma bond.
- In the structure, we have:
- C-H bonds: 6 (each C is bonded to H)
- C-C bonds: 2 (one from the double bond and one from the single bond)

Total sigma bonds:
- 6 (C-H) + 2 (C-C) = 8 sigma bonds.

### Step 4: Count the Pi Bonds
- A double bond consists of one sigma bond and one pi bond.
- In the structure, we have one double bond (C=C).

Total pi bonds:
- 1 (from the double bond).

### Step 5: Count the Lone Pairs of Electrons
- The oxygen atom in the hydroxyl group (–OH) has 6 valence electrons.
- It forms one bond with hydrogen and one bond with carbon, leaving it with 2 lone pairs.

Total lone pairs:
- 2 lone pairs on the oxygen atom.

### Summary of Counts
- **Sigma Bonds**: 8
- **Pi Bonds**: 1
- **Lone Pairs**: 2

### Final Answer
The enolic form of acetone contains:
- 8 sigma bonds
- 1 pi bond
- 2 lone pairs of electrons